Algebra
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Question
Function
Evaluate the derivative
Find the domain
Find the x-intercept/zero
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y′=−2x−18x2
Evaluate
y=−x2−6x3
Take the derivative of both sides
y′=dxd(−x2−6x3)
Use differentiation rule dxd(f(x)±g(x))=dxd(f(x))±dxd(g(x))
y′=−dxd(x2)−dxd(6x3)
Use dxdxn=nxn−1 to find derivative
y′=−2x−dxd(6x3)
Solution
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Calculate
dxd(6x3)
Use differentiation rule dxd(cf(x))=c×dxd(f(x))
6×dxd(x3)
Use dxdxn=nxn−1 to find derivative
6×3x2
Multiply the terms
18x2
y′=−2x−18x2
Show Solution
Testing for symmetry
Testing for symmetry about the origin
Testing for symmetry about the x-axis
Testing for symmetry about the y-axis
Not symmetry with respect to the origin
Evaluate
y=−x2−6x3
To test if the graph of y=−x2−6x3 is symmetry with respect to the origin,substitute -x for x and -y for y
−y=−(−x)2−6(−x)3
Simplify
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Evaluate
−(−x)2−6(−x)3
Multiply the terms
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Evaluate
6(−x)3
Rewrite the expression
6(−x3)
Multiply the numbers
−6x3
−(−x)2−(−6x3)
Rewrite the expression
−(−x)2+6x3
Rewrite the expression
−x2+6x3
−y=−x2+6x3
Change the signs both sides
y=x2−6x3
Solution
Not symmetry with respect to the origin
Show Solution
Rewrite the equation
r=0r=12cos3(θ)−cos2(θ)+cos(θ)(cos(θ)−24sin(θ))×∣cos(θ)∣r=−12cos3(θ)cos2(θ)+cos(θ)(cos(θ)−24sin(θ))×∣cos(θ)∣
Evaluate
y=−x2−6x3
Move the expression to the left side
y+x2+6x3=0
To convert the equation to polar coordinates,substitute x for rcos(θ) and y for rsin(θ)
sin(θ)×r+(cos(θ)×r)2+6(cos(θ)×r)3=0
Factor the expression
6cos3(θ)×r3+cos2(θ)×r2+sin(θ)×r=0
Factor the expression
r(6cos3(θ)×r2+cos2(θ)×r+sin(θ))=0
When the product of factors equals 0,at least one factor is 0
r=06cos3(θ)×r2+cos2(θ)×r+sin(θ)=0
Solution
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Factor the expression
6cos3(θ)×r2+cos2(θ)×r+sin(θ)=0
Solve using the quadratic formula
r=12cos3(θ)−cos2(θ)±(cos2(θ))2−4×6cos3(θ)sin(θ)
Simplify
r=12cos3(θ)−cos2(θ)±cos4(θ)−24sin(θ)cos3(θ)
Separate the equation into 2 possible cases
r=12cos3(θ)−cos2(θ)+cos4(θ)−24sin(θ)cos3(θ)r=12cos3(θ)−cos2(θ)−cos4(θ)−24sin(θ)cos3(θ)
Evaluate
r=12cos3(θ)−cos2(θ)+cos(θ)(cos(θ)−24sin(θ))×∣cos(θ)∣r=12cos3(θ)−cos2(θ)−cos4(θ)−24sin(θ)cos3(θ)
Evaluate
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Evaluate
12cos3(θ)−cos2(θ)−cos4(θ)−24sin(θ)cos3(θ)
Simplify the root
12cos3(θ)−cos2(θ)−cos(θ)(cos(θ)−24sin(θ))×∣cos(θ)∣
Use b−a=−ba=−ba to rewrite the fraction
−12cos3(θ)cos2(θ)+cos(θ)(cos(θ)−24sin(θ))×∣cos(θ)∣
r=12cos3(θ)−cos2(θ)+cos(θ)(cos(θ)−24sin(θ))×∣cos(θ)∣r=−12cos3(θ)cos2(θ)+cos(θ)(cos(θ)−24sin(θ))×∣cos(θ)∣
r=0r=12cos3(θ)−cos2(θ)+cos(θ)(cos(θ)−24sin(θ))×∣cos(θ)∣r=−12cos3(θ)cos2(θ)+cos(θ)(cos(θ)−24sin(θ))×∣cos(θ)∣
Show Solution
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